Question 682794
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{x\ -\ 1}\ -\ \frac{8}{x\ +\ 2}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{3}{x\ -\ 1}\right)\left(\frac{x\ +\ 2}{x\ +\ 2}\right)\ -\ \left(\frac{8}{x\ +\ 2}\right)\left(\frac{x\ -\ 1}{x\ -\ 1}\right)\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3(x\ +\ 2)\ -\ 8(x\ -\ 1)}{(x\ -\ 1)(x\ +\ 2)}\ =\ 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -5x\ +\ 14\ =\ x^2\ +\ x\ -\ 2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 6x\ -\ 16\ =\ 0]


Solve the quadratic in *[tex \LARGE x].  Check both roots to verify that both are elements of the solution set of the original equation. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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